pub fn dp_rec_mc(amount: u32) -> u32 {
    //实现最⼩找零纸币数
    //显然dp
    //dp[i] = min(dp[i], dp[i - coin] + 1)
    //硬币种类：[1, 2, 5, 10, 20, 50, 100]
    let coins = vec![1, 2, 5, 10, 20, 50, 100];
    let mut dp = vec![amount + 1; (amount + 1) as usize];
    dp[0] = 0;
    for i in 1..=amount {
        for &coin in coins.iter() {
            if i >= coin {
                dp[i as usize] = dp[i as usize].min(dp[(i - coin) as usize] + 1);
            }
        }
    }
    dp[amount as usize]
}
